Enthalpy in Extensive Units
H = U + pV
thermodynamics/what-is-energy-physics/what-is-enthalpy/”>Enthalpy is an extensive quantity, it depends on the size of the system, or on the amount of substance it contains. The SI unit of enthalpy is the joule (J). It is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. It is the thermodynamic quantity equivalent to the total heat content of a system.
On the other hand, energy can be stored in the chemical bonds between the atoms that make up the molecules. This energy storage on the atomic level includes energy associated with electron orbital states, nuclear spin, and binding forces in the nucleus.
Enthalpy is represented by the symbol H, and the change in enthalpy (delta H) in a process is H2 – H1.
There are expressions in terms of more familiar variables such as thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature andthermodynamics/thermodynamic-properties/what-is-pressure-physics/”> pressure:
dH = CpdT + V(1-αT)dp
Where Cp is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:
dH = CpdT
Example: Frictionless Piston – Heat – Enthalpy
A frictionless piston is used to provide a constant pressure of 500 kPa in a cylinder containing steam (superheated steam) of a volume of 2 m3 at 500 K. Calculate the final temperature, if 3000 kJ of heat is added.
Solution:
Using thermodynamics/steam-tables/”>steam tables we know, that the specific enthalpy of such steam (500 kPa; 500 K) is about 2912 kJ/kg. Since at this condition the steam has density of 2.2 kg/m3, then we know there is about 4.4 kg of steam in the piston at enthalpy of 2912 kJ/kg x 4.4 kg = 12812 kJ.
When we use simply Q = H2 − H1, then the resulting enthalpy of steam will be:
H2 = H1 + Q = 15812 kJ
From steam tables, such superheated steam (15812/4.4 = 3593 kJ/kg) will have a temperature of 828 K (555°C). Since at this enthalpy the steam have density of 1.31 kg/m3, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m3 x 1.67 = 3.34 m3 and ∆V = 3.34 m3 – 2 m3 = 1.34 m3.
The p∆V part of enthalpy, i.e. the work done is:
W = p∆V = 500 000 Pa x 1.34 m3 = 670 kJ
Enthalpy in Intensive Units – Specific Enthalpy
The enthalpy can be made into an intensive, or specific, variable by dividing by thethermodynamics/thermodynamic-properties/what-is-mass-and-weight/what-is-mass/”> mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. The specific enthalpy (h) of a substance is its enthalpy per unit mass. It equals to the total enthalpy (H) divided by the total mass (m).
h = H/m
where:
h = specific enthalpy (J/kg)
H = enthalpy (J)
m = mass (kg)
Note that the enthalpy is the thermodynamic quantity equivalent to the total heat content of a system. The specific enthalpy is equal to the thermodynamics/what-is-energy-physics/internal-energy-thermal-energy/specific-internal-energy/”>specific internal energy of the system plus the product of thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure and thermodynamics/thermodynamic-properties/what-is-specific-volume/”>specific volume.
h = u + pv
In general, enthalpy is a property of a substance, like pressure, temperature, and volume, but it cannot be measured directly. Normally, the enthalpy of a substance is given with respect to some reference value. For example, the specific enthalpy of water or steam is given using the reference that the specific enthalpy of water is zero at 0.01°C and thermodynamics/thermodynamic-properties/what-is-pressure-physics/atmospheric-pressure/”>normal atmospheric pressure, where hL = 0.00 kJ/kg. The fact that the absolute value of specific enthalpy is unknown is not a problem, however, because it is the change in specific enthalpy (∆h) and not the absolute value that is important in practical problems.
Calculate the amount of primary coolant, which is required to evaporate 1 kg of feedwater in a typical steam generator. Assume that there are no energy losses, this is only idealized example.
Balance of the primary circuit
The hot primary coolant (water 330°C; 626°F; 16MPa) is pumped into the steam generator through primary inlet. The primary coolant leaves (water 295°C; 563°F; 16MPa) the steam generator through primary outlet.
hI, inlet = 1516 kJ/kg
=> ΔhI = -206 kJ/kg
hI, outlet = 1310 kJ/kg
Balance of the feedwater
The feedwater (water 230°C; 446°F; 6,5MPa) is pumped into the steam generator through the feedwater inlet. The feedwater (secondary circuit) is heated from ~230°C 446°F to the boiling point of that fluid (280°C; 536°F; 6,5MPa). Feedwater is then evaporated and the pressurized steam (saturated steam 280°C; 536°F; 6,5 MPa) leaves the steam generator through steam outlet and continues to the steam turbine.
hII, inlet = 991 kJ/kg
=> ΔhII = 1789 kJ/kg
hII, outlet = 2780 kJ/kg
Balance of the steam generator
Since the difference in specific enthalpies is less for primary coolant than for feedwater, it is obvious that the amount of primary coolant will be higher than 1kg. To produce of 1 kg of saturated steam from feedwater, about 1789/206 x 1 kg = 8.68 kg of primary coolant is required.
Enthalpy Changes in Chemical Reactions
Consider the combustion of hydrogen in air. In a flame of pure hydrogen gas, burning in air, the hydrogen (H2) reacts with oxygen (O2) to form water (H2O) and releases energy.
Energetically, the process can be considered to require the energy to dissociate the H2 and O2, but then the bonding of the H2O returns the system to a bound state with negative potential. It is actually more negative than the bound states of the reactants, and the formation of the two water molecules is therefore an exothermic reaction, which releases 5.7 eV of energy. In words of enthalpy, the enthalpy of combustion is −286 kJ/mol:
2H2(g) + O2(g) → 2H2O(g)
In words of enthalpy, the enthalpy of combustion is −286 kJ/mol (energy per mol of molecular hydrogen):
2H2(g) + O2(g) → 2H2O(l) +572 kJ
The balance of energy before and after the reaction can be illustrated schematically with the state in which all atoms are free taken as the reference for energy.
Enthalpy of Vaporization
In general, when a material changes phase from solid to liquid, or from liquid to gas a certain amount of energy is involved in this change of phase. In case of liquid to gas phase change, this amount of energy is known as the enthalpy of vaporization, (symbol ∆Hvap; unit: J) also known as the (latent) heat of vaporization or heat of evaporation. Latent heat is the amount of heat added to or removed from a substance to produce a change in phase. This energy breaks down the intermolecular attractive forces, and also must provide the energy necessary to expand the gas (the pΔV work). When latent heat is added, no temperature change occurs. The enthalpy of vaporization is a function of the pressure at which that transformation takes place.
Latent heat of vaporization – water at 0.1 MPa (atmospheric pressure)
hlg = 2257 kJ/kg
Latent heat of vaporization – water at 3 MPa (pressure inside a steam generator)
hlg = 1795 kJ/kg
Latent heat of vaporization – water at 16 MPa (pressure inside a pressurizer)
hlg = 931 kJ/kg
The heat of vaporization diminishes with increasing pressure, while the boiling point increases. It vanishes completely at a certain point called the critical point. Above the critical point, the liquid and vapor phases are indistinguishable, and the substance is called a supercritical fluid.
The heat of vaporization is the heat required to completely vaporize a unit of saturated liquid (or condense a unit mass of saturated vapor) and it equal to hlg = hg − hl.
The heat that is necessary to melt (or freeze) a unit mass at the substance at constant pressure is the heat of fusion and is equal to hsl = hl − hs, where hs is the enthalpy of saturated solid and hl is the enthalpy of saturated liquid.
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